Derivation of 4D Lorentz Transformation

APPENDIX B

Derivation of 4D Lorentz Transformation Lorentz transformations (LT) have some important properties which we should explore before finding their generalization to Equation (6.16). All these properties are consequences of their defining equation:  ΛηΛ = η. (B.1)

B.1

GENERAL PROPERTIES

The first property is that the unit matrix is a LT, because it satisfies (B.1) trivially. The second property is that the product of two LTs is also a LT. Let Λ1 and Λ2 be two LTs. This means that 1 ηΛ1 = η and Λ 2 ηΛ2 = η. Λ Then    Λ 1 2 ηΛ1 2 = Λ2 Λ1 ηΛ1 Λ2 = Λ2 ηΛ2 = η.    =η

This shows that Λ1 Λ2 is a Lorentz transformation. The third property is that every LT has an inverse, which is also a LT. To show this, take the determinant of both sides of (B.1) to get       det η det Λ = det η ⇒ − det Λ  det Λ = −1 det Λ or (det Λ)2 = 1, because a matrix and its transpose have the same determinant. Therefore, det Λ = ±1, and since its determinant is not zero, Λ must have an  −1 = (Λ)  −1 inverse Λ−1 . Multiply Equation (B.1) by Λ−1 on the right and by Λ on the left to get −1   −1 ηΛ−1 ⇒ η = Λ −1 ηΛ−1 ,  −1 ΛηΛΛ  (Λ) =Λ showing that Λ−1 is a LT.1 Problem B.1 asks you to prove that the inverse of the transpose of a matrix is the transpose of its inverse. 1

These three properties turn the set of all Lorentz transformations into a group, the Lorentz group. But you don’t need to know anything about group theory to follow the rest of this appendix.

Special Relativity. DOI:10.1016/B978-0-12-810411-8.00021-3 Copyright © 2017 Elsevier Inc. All rights reserved.

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336 AP P EN D IX B: Derivation of 4D Lorentz Transformation

Lorentz transformations have an important subset. Consider the set of 4 × 4 matrices of the form ⎞ ⎛ 1 0 0 0   ⎜0 a11 a12 a13 ⎟ 1  0 ⎟ ⎜ ≡ , Λ≡⎝ 0 a21 a22 a23 ⎠ 0 A 0 a31 a32 a33 where on the right-hand side the 4 × 4 matrix is written in block diagonal form in which 0 is a column vector and A is the 3 × 3 matrix whose elements are given on the left-hand side. Write the matrix η also in block diagonal form and note that     1  0 1  0 1  0  ΛηΛ = 0  A 0 −1 0 A       1  0 1 0 1  0 = = . 0 − A 0 A 0 − AA The right-hand side becomes η if  AA = 1. You may recall that a matrix satisfying  AA = 1 is called orthogonal. Orthogonal 3 × 3 matrices are essentially rotations. Thus, three-dimensional rotations are special kinds of LTs.

B.2 DERIVATION OF THE LORENTZ BOOST Now we have all the machinery we need to obtain the most general LT, i.e., one for which the relative velocity β is along an arbitrary direction. All we need to do is use the freedom in choosing the orientation of the axes of our coordinate system: find a rotation which takes our coordinate system CS1 to another coordinate system CS2 in which β has the desired direction. First note that a rotation of the coordinate system is completely equivalent to the rotation of vectors in the opposite direction. Figure B.1 illustrates that. So, going from CS1 to CS2 is equivalent to rotating the vectors by some rotation matrix Λrot . Next let’s consider Equation (6.13), ⎞⎛ ⎞ ⎛ ⎞ ⎛ γ γβ 0 0 ct ct ⎜ x  ⎟ ⎜γβ γ 0 0⎟ ⎜ x ⎟ ⎟ ⎜ ⎟ or r = Λ0 r0 , ⎜ ⎟=⎜ 0 ⎝y ⎠ ⎝ 0 0 1 0⎠ ⎝ y ⎠ 0 0 0 1 z z   

(B.2)

Call this matrix Λ0 .

where the zero subscript indicates the initial (unrotated) quantities. Now multiply both sides of (B.2) by Λrot to get Λrot r0 = Λrot Λ0 r0

or Λrot r0 = Λrot Λ0 Λ−1 rot Λrot r0 .

Since r ≡ Λrot r0 is the new r 4-vector and r ≡ Λrot r0 is the new r 4-vector,  Λrot Λ0 Λ−1 rot must be the new Lorentz transformation Λ connecting r and r .

B.2 Derivation of the Lorentz Boost 337

FIGURE B.1 If you rotate CS1 you get CS2 . The (old) vectors have the same components in CS2 as the rotated (in the inverse direction) vectors would have in CS1 .

The only thing left to do is to construct Λ, for which purpose we only need Λrot .  Our Let βˆ = (βˆx , βˆy , βˆz ) denote the unit 3-vector in the direction of β. task, therefore, is to find a rotation which takes the three vector (1, 0, 0) to (βˆx , βˆy , βˆz ). Let A, with elements aij , be the 3 × 3 matrix that does the job. Namely, let ⎞⎛ ⎞ ⎛ ⎞ ⎛ βˆx a11 a12 a13 1 ⎝βˆy ⎠ = ⎝a21 a22 a23 ⎠ ⎝0⎠ . a31 a32 a33 0 βˆz Multiplying out the right-hand side shows that a11 = βˆx , a21 = βˆy , and a31 = βˆz . Therefore, ⎛ ⎞ βˆx a12 a13 A = ⎝βˆy a22 a23 ⎠ . βˆz a32 a33 If this is to be a rotation, it should satisfy  AA = A A = 1. You should convince yourself that if this holds and we consider the rows and columns of A as vectors, then each row (or each column) has a unit length and any two different rows (or different columns) are perpendicular to each other (see Problem B.2). The sought-after rotation has now been reduced to ⎞ ⎛ 1 0 0 0   ⎜0 βˆx a12 a13 ⎟ 1  0 ⎟ ⎜ ≡ . (B.3) Λrot = ⎝ 0 A 0 βˆy a22 a23 ⎠ 0 βˆz a32 a33 We also need the inverse of Λrot . Since Λrot is in block-diagonal form, its inverse is simply a block-diagonal matrix whose submatrices are inverses of the

338 AP P EN D IX B: Derivation of 4D Lorentz Transformation

corresponding submatrices of Λrot :

Λ−1 rot =



 1 0 0 A−1



⎛ 1  ⎜0 1  0 = =⎜ ⎝0 0  A 0 

0 βˆx a12 a13

0 βˆy a22 a23

⎞ 0 βˆz ⎟ ⎟. a32 ⎠ a33

(B.4)

Now we can put everything together and find Λ: ⎛

1 0 0 ⎜0 βˆx a12 Λ=⎜ ⎝0 βˆy a22 0 βˆz a32 ⎛ γ γβ ⎜γβx γ βˆx =⎜ ⎝γβy γ βˆy γβz γ βˆz

The most general Lorentz boost.

⎞⎛ ⎞⎛ 0 1 0 γ γβ 0 0 ⎜γβ γ 0 0⎟ ⎜0 βˆx a13 ⎟ ⎟⎜ ⎟⎜ 0 1 0⎠ ⎝0 a12 a23 ⎠ ⎝ 0 0 0 0 1 0 a13 a33 ⎞⎛ ⎞ 0 0 1 0 0 0 ⎜ ˆ ˆ ˆ ⎟ a12 a13 ⎟ ⎟ ⎜0 βx βy βz ⎟ a22 a33 ⎠ ⎝0 a12 a22 a32 ⎠ 0 a13 a23 a33 a32 a33

⎞ 0 βˆz ⎟ ⎟ a32 ⎠ a33

0 βˆy a22 a23

because β βˆx = βx etc. You should do the rest of matrix multiplications and use Problem B.2 to come up with the final result: ⎛

γ ⎜γβx ⎜ Λ=⎜ ⎝γβy γβz

γβx 1 + βˆx2 (γ − 1) βˆx βˆy (γ − 1)

γβy βˆx βˆy (γ − 1) 1 + βˆy2 (γ − 1)

γβz βˆx βˆz (γ − 1) βˆy βˆz (γ − 1)

βˆx βˆz (γ − 1)

βˆy βˆz (γ − 1)

1 + βˆz2 (γ − 1)

⎞ ⎟ ⎟ ⎟. ⎠

(B.5)

Equation (B.5) does not give the most general LT matrix, because it is fully described by the relative velocity β of two observers. As mentioned before the full set of Lorentz transformations include rotations as well. Equation (B.5) gives  Lorentz only a so-called Lorentz boost, which we sometimes write as Λ(β).  boosts have the additional property that Λ = Λ, which is not shared by a general Lorentz transformation.

B.2.1

Transformation of Position and Time

Now that we have Λ, let’s apply it to the new r ≡ (ct, x, y, z). Calling the result of this application r , and using matrices, you should be able to get ⎞ ⎛ ⎞ γ ct + γ β · r ct  ⎜ x  ⎟ ⎜x + βx γ ct + βˆx (γ − 1)βˆ · r⎟ ⎜ ⎟=⎜ ⎟ ⎝ y ⎠ ⎝y + βy γ ct + βˆy (γ − 1)βˆ · r⎠ , z z + βz γ ct + βˆz (γ − 1)βˆ · r ⎛

(B.6)

B.2 Derivation of the Lorentz Boost 339

where all components are relative to the new coordinate system. This shows that ct  = γ (ct + β · r) and ⎛ ⎞ ⎛ ⎞ x + βx γ ct + βˆx (γ − 1)βˆ · r x ⎝y  ⎠ = ⎝y + βy γ ct + βˆy (γ − 1)βˆ · r⎠ z z + βz γ ct + βˆz (γ − 1)βˆ · r ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ βˆx x βx = ⎝y ⎠ + γ ct ⎝βy ⎠ + (γ − 1)βˆ · r ⎝βˆy ⎠ , βz z βˆz or in vector notation 1  − 1)β · r. β(γ β2

ˆ − 1)βˆ · r = r + γ ct β + r  = r + γ ct β + β(γ

With β 2 = (γ 2 − 1)/γ 2 , the final result, which is stated in Equation (6.16), is trivially reached.

B.2.2

The 3 × 3 Submatrix

It is sometimes useful to separate the 3 × 3 matrix in Λ, which is obtained by eliminating the first row and column. So, first write Λ in block form:    γ γ β (B.7) Λ= ↔ , γ β Λ ↔ where Λ is the 3 × 3 submatrix of Λ. Then, with the help of (B.5), write ⎛ ⎞ 1 + βˆx2 (γ − 1) βˆx βˆy (γ − 1) βˆx βˆz (γ − 1) ↔ ⎜ ˆ ˆ ⎟ Λ = ⎝ βx βy (γ − 1) 1 + βˆy2 (γ − 1) βˆy βˆz (γ − 1)⎠ , βˆy βˆz (γ − 1) 1 + βˆz2 βˆx βˆz (γ − 1) or ⎛

1 ↔ ⎝ Λ= 0 0

0 1 0

⎛ 2 ⎞ βˆx 0 ⎜ˆ ˆ ⎠ 0 + (γ − 1) ⎝βx βy 1 βˆ βˆ x z

βˆx βˆy βˆy2 βˆy βˆz

⎞ βˆx βˆz ⎟ βˆy βˆz ⎠ .

This can also be expressed in indexed form as ↔ Λ ij = δij + (γ − 1)βˆi βˆj , where δij is the Kronecker delta defined by  1 if i = j δij = 0 if i =  j.

(B.8)

βˆz2

(B.9)

(B.10)

340 AP P EN D IX B: Derivation of 4D Lorentz Transformation

B.2.3

Inverse of a Boost

The inverse Lorentz boost is also important, because sometimes it is desired to go from the new RF to the old one. This transformation is accomplished by the inverse of the 4 × 4 matrix (B.5). Finding Λ−1 may appear daunting, but physics makes it trivial: the only difference between going from old to new and going from new to old is the direction of velocity. Therefore, changing β to −β should produce the inverse: ⎞ ⎛ −γβy −γβz γ −γβx ⎜−γβx 1 + βˆx2 (γ − 1) βˆx βˆy (γ − 1) βˆx βˆz (γ − 1) ⎟ ⎟ ⎜ −1 (B.11) Λ =⎜ ⎟. 2 ⎝−γβy βˆx βˆy (γ − 1) 1 + βˆy (γ − 1) βˆy βˆz (γ − 1) ⎠ −γβz βˆy βˆz (γ − 1) 1 + βˆz2 (γ − 1) βˆx βˆz (γ − 1) You should check that Λ−1 Λ = ΛΛ−1 = 1. If you multiply both sides of Equation (B.1) on the right by Λ−1 η, you get =η2 =1

    = ηΛ−1 η,  η ΛΛ−1 η = ηΛ−1 η ⇒ Λ Λ    =1

 = Λ) leads to which in the case of Lorentz boosts (for which Λ  = ηΛ(−β)η   = ηΛ(β)η.  Λ(β) or Λ(−β)

(B.12)

These are useful relations which are used occasionally.

B.3 PROBLEMS B.1. Take the transpose of A−1 A = AA−1 = 1 to show that the inverse of the transpose is the transpose of the inverse. You need both matrix products because an inverse must be a left inverse as well as a right inverse. B.2. Show that  AA = A A = 1 for a 3 × 3 matrix implies that each row (or column) of A has a unit length and the dot product of any two different rows (or different columns) vanishes. B.3. By writing out the 4 × 4 matrices, verify that Λ−1 rot as given by Equation (B.4) is the inverse of Λrot as given by Equation (B.3). 2 + a2 = 1 − B.4. Using relations such as βˆx βˆy + a12 a22 + a13 a23 = 0 and a12 13 2 (βˆx ) , obtained from Problem B.2, derive Equation (B.5).

B.5. Multiply the matrix of Equation (B.5) by the column 4-vector of r to obtain Equation (B.6). B.6. Show directly that the matrices of (B.5) and (B.11) satisfy Λ−1 Λ = ΛΛ−1 = 1.